Ultimate Ruin Calculation Workshop

Purpose and Model

An ultimate ruin calculation asks whether the surplus process ever falls below zero, not merely whether ruin occurs during a specified period. The calculation therefore depends on both the claim process and the premium income accumulated over an unlimited horizon. A sound solution begins by checking the model and the net profit condition, then choosing a result that applies to the claim severity distribution.

Throughout this tutorial, use the classical compound Poisson model. The insurer begins with initial surplus $u\geq 0$, receives premium continuously at rate $c>0$, and pays claims immediately. Claim arrivals follow a Poisson process $N(t)$ with rate $\lambda>0$ per year. Claim amounts $X_1,X_2,\ldots$ are independent and identically distributed positive random variables, independent of $N(t)$. There is no interest, expense, or other cash flow in the model.

Definition: The surplus at time $t$ is $U(t)=u+ct-\sum_{j=1}^{N(t)}X_j$. The ultimate ruin time is $\tau=\inf\{t\geq 0:U(t)<0\}$, and the ultimate ruin probability is $\psi(u)=P(\tau<\infty)$. For continuous claim amounts, replacing the strict inequality by a non-strict one does not change the probability.

The expected claim cost per year is $\lambda E[X]$. The net profit condition is $c>\lambda E[X]$: premium income must exceed expected claim outgo. It is necessary for the usual positive safety-loading results and positive adjustment-coefficient bounds. If $c<\lambda E[X]$, the surplus has negative long-run drift, and under the stated model ruin occurs eventually with probability one. At equality, the usual positive-loading formulas do not apply; for a nondegenerate claim process with positive arrival rate, ultimate ruin also occurs with probability one.

\frac{U(t)}{t}\ \longrightarrow\ c-\lambda E[X]\quad\text{almost surely as }t\to\infty

This long-run drift explains why the net profit check comes first. It does not give the probability of ruin when the drift is positive; that probability depends on the size and tail of claims as well as on $u$.

Choosing an Ultimate-Ruin Method

Two useful tools are an exact formula for exponential claims and the Lundberg bound. An exact formula gives $\psi(u)$ under its specified severity assumptions. A bound gives a guaranteed upper limit, not generally the exact probability.

Available information	Method	What the result means
Exponential claim amounts and positive loading	Exact exponential-claim formula	The model's ultimate ruin probability
A moment-generating function and an admissible positive $r$	Lundberg bound	An upper bound on ultimate ruin probability
Net profit condition fails	Do not use a positive-loading formula or bound	Under the stated conditions, ruin is certain eventually

Write $M_X(r)=E[e^{rX}]$ when this moment-generating function is finite. A positive number $r$ is admissible for the Lundberg bound if $\lambda(M_X(r)-1)\leq cr$. In particular, a positive solution $R$ of $\lambda(M_X(R)-1)=cR$ gives the familiar adjustment-coefficient bound when it exists. An admissible $r$ need not be the root; the bound remains valid, though it may be less sharp.

\psi(u)\leq e^{-ru} \quad\text{if}\quad r>0,\qquad \lambda\bigl(M_X(r)-1\bigr)\leq cr

The bound can be justified by considering $e^{-rU(t)}$. Its expected multiplicative change per unit time is governed by $\lambda(M_X(r)-1)-cr$, which is nonpositive for an admissible $r$. Thus it is a nonnegative supermartingale after scaling by its initial value. If ruin occurs, the surplus has crossed below zero, so $e^{-rU(t)}$ has reached at least 1. The maximal inequality for nonnegative supermartingales then gives the stated upper bound $e^{-ru}$.

Exact Formula for Exponential Claims

Suppose $X$ is exponential with mean $\beta>0$, so $E[X]=\beta$, and suppose the net profit condition holds. Define $\rho=\lambda\beta/c$, the ratio of expected annual claim cost to premium income. Then $0<\rho<1$, and the exact ultimate ruin probability is

\psi(u)=\rho e^{-Ru}, \qquad R=\frac{1-\rho}{\beta} =\frac{c-\lambda\beta}{c\beta}.

Here $R$ has units inverse to the units of claim amount, so $Ru$ is dimensionless. The adjustment coefficient is the positive solution of the Lundberg equation: for exponential claims, $M_X(r)=1/(1-\beta r)$ when $0\leq r<1/\beta$. Substitution in $\lambda(M_X(r)-1)=cr$ gives the positive root $r=(c-\lambda\beta)/(c\beta)$.

Why the exact formula holds: For exponential claims, the ultimate ruin probability satisfies the renewal equation $c\psi'(u)=\lambda\psi(u)-\lambda\int_0^u\psi(u-x)f_X(x)\,dx-\lambda S_X(u)$, with boundary value $\psi(0)=\lambda E[X]/c=\rho$. Try $\psi(u)=C e^{-Ru}$, where $R=(1-\rho)/\beta$. The convolution in the equation is

\int_0^u C e^{-R(u-x)}\frac{1}{\beta}e^{-x/\beta}\,dx =\frac{C}{\rho}\left(e^{-Ru}-e^{-u/\beta}\right).

Using $S_X(u)=e^{-u/\beta}$, the coefficient of $e^{-u/\beta}$ on the right side of the renewal equation is zero when $C=\rho$. The remaining coefficient is $-\lambda(1-\rho)$, which equals the coefficient on the left because $cR\rho=c(1-\rho)\rho/\beta=\lambda(1-\rho)$. The candidate therefore satisfies the equation and boundary value, giving the stated solution. This derivation depends on exponential severity; it is not an exact formula for arbitrary claim distributions.

Worked Example: Exact Ruin Probability and a Bound

A hypothetical insurer has Poisson claim arrivals at rate $\lambda=1.5$ per year. Individual claim amounts are exponential with mean $\beta=\$800$. The continuous premium rate is $c=\$1{,}440$ per year, and initial surplus is $u=\$1{,}200$. Find the ultimate ruin probability and compare it with the adjustment-coefficient bound.

Method and assumptions. The model is classical, claims are independent exponential amounts, and there is no interest. First check the net profit condition: $\lambda E[X]=1.5(\$800)=\$1{,}200$ per year, while $c=\$1{,}440$ per year. Since $\$1{,}440>\$1{,}200$, the condition holds.

The claim-cost ratio is $\rho=\lambda\beta/c=\$1{,}200/\$1{,}440=5/6$. The adjustment coefficient is $R=(1-\rho)/\beta=(1/6)/\$800=1/4{,}800$ per dollar. Thus $Ru=\$1{,}200/\$4{,}800=0.25$.

Apply the exact formula: $\psi(u)=(5/6)e^{-0.25}=(5/6)(0.778800783\ldots)=0.649000653\ldots$. To four decimal places, the ultimate ruin probability is 0.6490.

The Lundberg bound uses the same root $R$: $e^{-Ru}=e^{-0.25}=0.778800783\ldots$, or 0.7788 to four decimal places. It is larger than the exact probability, as an upper bound may be. The check is $0.6490\leq0.7788$. The common mistake would be to report the bound as the exact probability; the exact formula includes the prefactor $\rho=5/6$.

Using a Bound When Severity Is Not Exponential

For a non-exponential claim distribution, the exponential exact formula does not apply. If the moment-generating function exists at a positive value $r$, the Lundberg bound can still be used after checking its inequality. Do not assume that any positive $r$ works: the premium term $cr$ must be at least as large as $\lambda(M_X(r)-1)$.

Worked Example: Testing a Proposed Lundberg Parameter

In a hypothetical model, claim arrivals are Poisson at rate $\lambda=2$ per year. Each claim is independently $\$500$ or $\$1{,}500$, with probability $1/2$ each. The premium rate is $c=\$2{,}300$ per year, initial surplus is $u=\$4{,}000$, and a proposed bound uses $r=0.0002$ per dollar. Determine whether the parameter is admissible and, if so, calculate the bound.

Method and common mistake avoided. Check the net profit condition, evaluate the moment-generating function using the dollar claim amounts, and then compare both sides of the admissibility inequality. The mean claim amount is $E[X]=0.5(\$500)+0.5(\$1{,}500)=\$1{,}000$, so expected claim cost is $\lambda E[X]=\$2{,}000$ per year. The condition holds because $\$2{,}300>\$2{,}000$.

At the proposed $r$, $r(\$500)=0.1$ and $r(\$1{,}500)=0.3$. Therefore $M_X(r)=0.5e^{0.1}+0.5e^{0.3}$ $=0.5(1.105170186)+0.5(1.349858808)=1.227514497$. Hence $\lambda(M_X(r)-1)=2(0.227514497)=0.455028993$ per year, while $cr=(\$2{,}300)(0.0002\text{ per dollar})=0.46$ per year. Since $0.455028993\leq0.46$, $r$ is admissible.

The bound is $e^{-ru}=e^{-(0.0002)(\$4{,}000)}=e^{-0.8}=0.449328964\ldots$, or 0.4493 to four decimal places. This is an upper bound, not an exact ruin probability. The common mistake is to round the two sides before checking the inequality or to treat the proposed $r$ as a root without solving the Lundberg equation.

Interpreting a Target Ultimate Ruin Probability

An ultimate probability concerns the entire future lifetime of the surplus process. It is not interchangeable with a probability calculated over one year or another fixed period. When a question asks for the capital needed to meet an ultimate ruin target, use an ultimate formula or bound appropriate to the model, and solve for $u$ only after confirming that the selected result is applicable.

Worked Example: Capital Required for an Ultimate Ruin Target

A hypothetical insurer has Poisson claim arrivals at rate $\lambda=4$ per year and exponential claim amounts with mean $\beta=\$2{,}500$. Its premium rate is $c=\$11{,}000$ per year. Find the smallest initial surplus for which the exact ultimate ruin probability is at most 0.05.

Method and assumptions. Apply the exact exponential-claim formula. Expected annual claim cost is $\lambda\beta=4(\$2{,}500)=\$10{,}000$, so the net profit condition holds because $\$11{,}000>\$10{,}000$. Then $\rho=\$10{,}000/\$11{,}000=10/11$, and $R=(1-\rho)/\beta=(1/11)/\$2{,}500=1/(\$27{,}500)$.

The target requires $(10/11)e^{-u/27{,}500}\leq0.05$. Dividing by $10/11$ gives $e^{-u/27{,}500}\leq0.055=11/200$. Taking logarithms and reversing the inequality when multiplying by the negative denominator gives $u\geq27{,}500\ln(200/11)$.

Since $\ln(200/11)=2.900422094\ldots$, the required capital is $u\geq\$79{,}761.6076\ldots$. To the nearest cent, the smallest required initial surplus is \$79,761.61. Substitution at the unrounded threshold gives $\psi(u)=0.05$; any smaller surplus gives a larger ultimate ruin probability. A common mistake is to use a finite-horizon result for this target or to round the logarithm before computing the capital.

Checks Before Reporting an Answer

Write the model and units.
Identify the arrival rate per year, claim amount units, premium rate per year, and initial surplus. Keep $r$ in inverse claim-amount units so that $rX$ and $ru$ are dimensionless.

Check net profit.
Calculate $\lambda E[X]$ and verify $c>\lambda E[X]$ before using the positive-loading exact formula or Lundberg bound.

Choose an applicable result.
Use the exact exponential formula only for exponential claims. For a bound, verify $r>0$, finiteness of $M_X(r)$, and $\lambda(M_X(r)-1)\leq cr$.

Interpret the answer.
Label a result as exact or as an upper bound, and state that it is ultimate rather than finite-horizon ruin probability.

A frequent exam trap is to check only that the premium rate is positive. Positivity of $c$ is not the net profit condition: compare it with the expected annual claims $\lambda E[X]$. Another is a time-unit mismatch. If $\lambda$ is given per month but $c$ per year, convert one before comparing them. A numerically plausible ruin probability can still be invalid if the model assumptions or units do not match the formula.

Key takeaway: Ultimate ruin calculations are model-specific. Verify the net profit condition, use a formula or bound whose assumptions match the severity model, and distinguish an exact probability from an upper bound. A bound is useful for risk control, but it is not an estimate of equality.

Check Your Understanding

For each hypothetical classical-model problem, identify the applicable check or formula before computing. Probabilities are rounded to four decimal places.

Claims arrive at rate $1$ per year and have exponential amounts with mean $\$500$. The premium rate is $\$625$ per year and initial surplus is $\$1{,}000$. Find the exact ultimate ruin probability. (Answer: $0.5363$)
Claims arrive at rate $3$ per year, with mean amount $\$400$. The premium rate is $\$1{,}150$ per year. Does the net profit condition hold, and what is the ultimate ruin probability under the classical model? (Answer: The condition fails because $\$1{,}150<\$1{,}200$; $\psi(u)=1$ for finite $u$.)
Claims arrive at rate $1.5$ per year and are $\$200$ or $\$800$, each with probability $1/2$. The premium rate is $\$900$ per year, initial surplus is $\$2{,}000$, and $r=0.0005$ per dollar. Is the parameter admissible, and what bound does it give? (Answer: Yes; $\lambda(M_X(r)-1)=0.447746\ldots\leq cr=0.45$, and $\psi(u)\leq e^{-1}=0.3679$.)
Claims arrive at rate $2$ per year and are exponential with mean $\$1{,}000$. The premium rate is $\$2{,}500$ per year and initial surplus is $\$5{,}000$. Calculate the exact ultimate ruin probability. (Answer: $0.2943$)
Claims arrive at rate $0.8$ per year and are exponential with mean $\$1{,}200$. The premium rate is $\$1{,}000$ per year, and initial surplus is $\$2{,}400$. Check net profit, then calculate exact ultimate ruin probability if the formula applies. (Answer: The condition holds; $\rho=0.96$, $R=1/\$30{,}000$, and $\psi(u)=0.96e^{-0.08}=0.8862$.)

What You'll Learn